Answer:
8.1×10^-8 mols-1
Explanation:
Now we have the mass of copper sulphate produced after three days. Recall that the rate of reaction is given as;
Rate= change in the concentration of product/time
At the beginning of the reaction, there was 0 moles of copper sulphate
After 72 hours or 259200 seconds, there was 3.4g/160gmol-1 = 0.021 moles of copper sulphate.
Note that 160gmol-1 is the molar mass of copper sulphate.
Hence;
Rate of reaction= 0.021 moles /259200 seconds
Hence, the rate of reaction is 8.1×10^-8 mols-1
Rate of reaction= 8.1×10^-8 mols-1
Answer is: 31,45%.
mrs₁(C₉H₁₆O₄-<span>azelaic acid) = 12g.
mr</span>₂(C₉H₁₆O₄) = 50g.
ω₂(C₉H₁₆O₄) = 15% = 0,15.
mrs₂(C₉H₁₆O₄) = mr₂·ω₂ = 50g·0,15 = 7,5g.
mrs₃(C₉H₁₆O₄) = mrs₁ + mr₂ = 12g + 7,5g = 19,5g.
mr₃ = mr₂ + mr₂ = 50g + 12g = 62g.
ω₃ = mrs₃÷mr₃ = 19,5g ÷ 62g = 31,45% = 0,3145.
It is scandium or titanium iron chroniclemium vanadium manganese
Explanation:
The - 3 degree C( carbon atom) 2p atomic orbital + methyl C-H sigma molecular orbital because one C-H bond has to dissolve its bond and provide the H that is sigma molecular orbital and the carbonation is type 3 degree sp2 carbon.
Hyperconjugation is the stabilizing effect arising from the electrons ' engagement in a π-bond (usually C-H or C-C) with a neighboring empty or partly filled p-orbital or π-orbital to provide an expanded molecular orbital that enhances system stability.
Answer : The final temperature would be, 791.1 K
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= 
= activation energy for the reaction = 265 kJ/mol = 265000 J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature = ?
Now put all the given values in this formula, we get:
![\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B4%5Ctimes%20K_1%7D%7BK_1%7D%29%3D%5Cfrac%7B265000J%2Fmol%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B733K%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
Therefore, the final temperature would be, 791.1 K
